The equation $y^{2} e^{x y} = 9 e^{- 3} . x^{2}$ defines $y$ as a differentiable function of x. The value of $\frac{d y}{d x}$ for $x = - 1$ and $y = 3$ is

- \frac{15}{2}

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- \frac{9}{5}

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3

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15

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Solution

The correct option is D $15$
$y^{2} e^{x y} = 9 e^{- 3} . x^{2}$
Differentiate both sides
$\frac{d}{d x} (y^{2} e^{x y}) = \frac{d}{d x} (9 e^{- 3} x^{2})$
$2 y \frac{d y}{d x} e^{x y} + y^{2} e^{x y} (x \frac{d y}{d x} + y) = 9 e^{- 3} (2 x)$
On solving, we get
$\frac{d y}{d x} = \frac{18 x e^{- 3} - y^{3} e^{x y}}{2 y e^{x y} + y^{2} e^{x y} x}$
putting $x = - 1, y = 3$ , we get
$\frac{d y}{d x} = \frac{- 18 e^{- 3} - 27 e^{- 3}}{6 e^{- 3} - 9 e^{- 3}} = \frac{- 45 e^{- 3}}{- 3 e^{- 3}} = 15$

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Similar questions

y^{2} e^{x y} = 9 e^{- 3} \cdot x^{2}

y

x .

\frac{d y}{d x}

x = - 1

y = 3

x^{2} = 1 + (\frac{x}{y})^{- 1} \frac{d y}{d x} + \frac{(\frac{x}{y})^{- 2} (\frac{d y}{d x})^{2}}{2!} + \frac{(\frac{x}{y})^{- 3} (\frac{d y}{d x})^{3}}{3!} + . . . .

(x^{3} + y^{3}) d y - x^{2} y d x = 0

y

x

\frac{d^{2} x}{d y^{2}} {(\frac{d y}{d x})}^{3} + \frac{d^{2} y}{d x^{2}} =

y

y^{3} \frac{d y}{d x} + x^{3} = 0, y (0) = 1,

y (- 1)

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